/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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NO

After renaming modulo the bijection { a ↦ 0, b ↦ 1, c ↦ 2 },
it remains to prove termination of the 2-rule system
{ 0 0 0 ⟶ 1 2 ,
  2 1 ⟶ 0 0 0 0 2 }

Loop of length 15 starting with a string of length 4
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.cb.bb
	rule cb-> aaaac at position 0
.aaaac.bb
	rule aaa-> bc at position 0
.bcac.bb
	rule cb-> aaaac at position 3
.bcaaaaac.b
	rule aaa-> bc at position 2
.bcbcaac.b
	rule cb-> aaaac at position 6
.bcbcaaaaaac.
	rule aaa-> bc at position 4
.bcbcbcaaac.
	rule cb-> aaaac at position 3
.bcbaaaaccaaac.
	rule aaa-> bc at position 3
.bcbbcaccaaac.
	rule aaa-> bc at position 8
.bcbbcaccbcc.
	rule cb-> aaaac at position 7
.bcbbcacaaaaccc.
	rule aaa-> bc at position 7
.bcbbcacbcaccc.
	rule cb-> aaaac at position 6
.bcbbcaaaaaccaccc.
	rule aaa-> bc at position 5
.bcbbcbcaaccaccc.
	rule cb-> aaaac at position 4
.bcbbaaaaccaaccaccc.
	rule aaa-> bc at position 4
.bcbbbcaccaaccaccc.